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Old   April 5, 2019, 16:05
Default Questions about mesh initialization in CONVERGE
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tmy
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Mingyuan Tao
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Hi all,

I have some questions about the way CONVERGE use to initialize mesh. I have a rectangular geometry with L=0.2m, and H=0.05m, and I am conducting mesh independent calculation currently. However, there are two problems:

1. I have the fix_embedding on with 20 layers and scales=2. If I set dx=6e-4, dy=8e-4, grid_scale=0, it gives a basically good fix_embedding on the boundary. However, if I set dx=1.2e-3, dy=1.6e-3, grid_scale=1, which in my understanding, should function the same as the previous setting, the generated mesh shows a much thinker fix embedding on the boundary, and therefore more cells are generated for calculation. I am really confused about the difference between these two settings.

2. Another confusing problem also arises from the two different dx, dy, grid_scale settings. When I use parallel calculation command:
"nohup mpirun -n 50 converge-2.4.man super >a.log &"
the memory usage by setting dx=6e-4, dy=8e-4, grid_scale=0 is much higher than the usage by setting dx=1.2e-3, dy=1.6e-3, grid_scale=1. Sometimes depending on how small dx, dy can be, the case with grid_scale=0 cannot even be calculated because almost all memory is used up.

Could anyone help me these two problems? I appreciate it!
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Old   April 5, 2019, 19:23
Default Re: Questions about mesh initialization in CONVERGE
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Vigneshwar Ravisankar
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Hi,

1. The embedded grid are generated based on the base grid value at each scale and in further calculation the grid scaling is performed. So in case of dx=6e-4, dy=8e-4, grid_scale=0, the base grid is considered as dx=6e-4, dy=8e-4. And further embedding is performed for scaling of 2. In case of dx=1.2e-3, dy=1.6e-3, grid_scale=1, The dx and dy at scale=0 is dx=1.2e-3, dy=1.6e-3. So this is used as base grid for initial embedding. Then for further calc, the base grid is re-scaled to dx=6e-4, dy=8e-4 and embedding is done to scale 2 with new base grid value.

2. The block sice for parallel calculation is calculated as:
Block size= dx * 2^PS, Where PS is parallel scaling.
Assuming PS is 1.
So for the first case, Block size is 1.2e-3 (=6e-4*2), 1.6e-3 (=8e-4*2). so for a domain of L=0.2m, and H=0.05m. # block will be close to 5208, with 4 grids in each. (roughly in simple terms)
For the 2nd case Block size is 2.4e-3 (=1.2e-3*2), 3.2e-3 (=1.6e-3*2) so for a domain of L=0.2m, and H=0.05m. # block will be close to 1302 (1/4th), with 4 grids in each. (roughly in simple terms).

So the first case using a lot more memory due to high # of blocks. Change the PS value accordingly for better load balance


Hope that's clear. Let me know if you still have trouble to understand my explanation.


Best regards,

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Vigneshwar Ravisankar
Application Engineer
CONVERGECFD
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Old   April 5, 2019, 20:19
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tmy
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Mingyuan Tao
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Thanks for your reply, that's clear!

Quote:
Originally Posted by rvigneshwar View Post
Hi,

1. The embedded grid are generated based on the base grid value at each scale and in further calculation the grid scaling is performed. So in case of dx=6e-4, dy=8e-4, grid_scale=0, the base grid is considered as dx=6e-4, dy=8e-4. And further embedding is performed for scaling of 2. In case of dx=1.2e-3, dy=1.6e-3, grid_scale=1, The dx and dy at scale=0 is dx=1.2e-3, dy=1.6e-3. So this is used as base grid for initial embedding. Then for further calc, the base grid is re-scaled to dx=6e-4, dy=8e-4 and embedding is done to scale 2 with new base grid value.

2. The block sice for parallel calculation is calculated as:
Block size= dx * 2^PS, Where PS is parallel scaling.
Assuming PS is 1.
So for the first case, Block size is 1.2e-3 (=6e-4*2), 1.6e-3 (=8e-4*2). so for a domain of L=0.2m, and H=0.05m. # block will be close to 5208, with 4 grids in each. (roughly in simple terms)
For the 2nd case Block size is 2.4e-3 (=1.2e-3*2), 3.2e-3 (=1.6e-3*2) so for a domain of L=0.2m, and H=0.05m. # block will be close to 1302 (1/4th), with 4 grids in each. (roughly in simple terms).

So the first case using a lot more memory due to high # of blocks. Change the PS value accordingly for better load balance


Hope that's clear. Let me know if you still have trouble to understand my explanation.


Best regards,

__________________
Vigneshwar Ravisankar
Application Engineer
CONVERGECFD
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